lecture10 - Physics 2102 Gabriela Gonzlez Any two charged conductors form a capacitor Capacitance C= Q\/V Simple Capacitors Parallel plates C = 0 A\/d

# lecture10 - Physics 2102 Gabriela Gonzlez Any two charged...

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1 Physics 2102 Gabriela González • Any two charged conductors form a capacitor. •Capacitance : C= Q/V •Simple Capacitors: Parallel plates : C = ε 0 A/d Spherical : C = ε 0 4 π ab/(b-a) Cylindrical : C = ε 0 2 π L/ln(b/a)
2 Start out with uncharged capacitor Transfer small amount of charge dq from one plate to the other until charge on each plate has magnitude Q How much work was needed? dq Energy stored in capacitor:U = Q 2 /(2C) = CV 2 /2 View the energy as stored in ELECTRIC FIELD For example, parallel plate capacitor: Energy DENSITY = energy/volume = u = volume = Ad General expression for any region with vacuum (or air)
3 If the space between capacitor plates is filled by a dielectric, the capacitance INCREASES by a factor κ This is a useful, working definition for dielectric constant. Typical values of κ : 10 - 200 +Q - Q DIELECTRIC C = κε 0 A/d Capacitor has charge Q, voltage V Battery remains connected while dielectric slab is inserted. Do the following increase, decrease or stay the same: Potential difference? – Capacitance? – Charge? Electric field? dielectric slab
4 Initial values: capacitance = C ; charge = Q ; potential difference = V ; electric field = E ; Battery remains connected V is FIXED; V new = V ( same ) C new = κ C ( increases ) Q new = ( κ C)V = κ Q ( increases ).

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• Spring '08
• GIMMNACO
• Physics, Capacitance, Charge, Potential difference, Electric charge, Energy density, Georg Simon Ohm

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